What is the time complexity of Remove Sub-Folders from the Filesystem?

The optimal solution has a time complexity of O(n*l).

What is the space complexity of Remove Sub-Folders from the Filesystem?

The space complexity is O(n*l).

What data structures are used to solve Remove Sub-Folders from the Filesystem?

This problem uses Array, String, Depth-First Search, Trie. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Remove Sub-Folders from the Filesystem?

Insight 1: Sorting the folder paths by length allows us to efficiently check for sub-folders. A shorter path can only be a parent of a longer path, not the other way around. Insight 2: Using a Trie data structure helps in quickly checking if a path is a sub-folder of any existing path.

Which companies ask LeetCode 1233. Remove Sub-Folders from the Filesystem in interviews?

This problem is frequently asked at Snowflake, Verkada.

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Remove Sub-Folders from the Filesystem - LeetCode 1233 Solution

Remove Sub-Folders from the Filesystem - Complete Solution Guide

Remove Sub-Folders from the Filesystem is LeetCode problem 1233, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a list of folders folder , return the folders after removing all sub-folders in those folders . You may return the answer in any order . If a folder[i] is located within another folder[j] , it is called a sub-folder of it. A sub-folder of folder[j] must start with folder[j] , followed by a "/" . For example, "/a/b" is a sub-folder of "/a" , but "/b" is not a sub-folder of "/a/b/c" . The format of a path is one or more concatenated strings of the form: '/' followed by one or more lowercase

Detailed Explanation

The problem asks us to remove sub-folders from a given list of folder paths. A sub-folder is defined as a folder whose path starts with another folder's path, followed by a '/'. For example, if we have '/a' and '/a/b', then '/a/b' is a sub-folder of '/a', and we should only keep '/a' in the final result. The input is a list of strings representing folder paths, and the output should be a list of strings containing only the parent folders (i.e., folders that are not sub-folders of any other folder in the input).

Solution Approach

The solution utilizes a Trie (prefix tree) data structure to store the folder paths and efficiently identify sub-folders. A Trie is built by inserting each folder path. Then, a Breadth-First Search (BFS) is used to traverse the Trie. While traversing, if a node represents the end of a folder path, it's considered a parent folder. Sub-folders are identified because once a parent folder is found during the BFS, we stop exploring further down that branch, effectively pruning the sub-folders.

Step-by-Step Algorithm

Step 1: Initialize an empty Trie (implemented using nested dictionaries or HashMaps).
Step 2: Iterate through the input `folder` list and insert each `path` into the Trie. Each level of the Trie represents a directory segment (e.g., 'a', 'b', 'c' in '/a/b/c'). A special marker ('$' in Python/Java or a boolean flag in C) is used to indicate the end of a path and store the full folder string at that node.
Step 3: Initialize an empty list `res` to store the parent folder paths.
Step 4: Perform a Breadth-First Search (BFS) on the Trie, starting from the root.
Step 5: In each BFS iteration, retrieve a node from the queue.
Step 6: Check if the current node contains the end-of-path marker. If it does, it means the node represents a parent folder.
Step 7: If the node represents a parent folder, add the corresponding folder path to the `res` list and skip exploring the children of that node. This effectively removes any sub-folders of the parent folder from consideration.
Step 8: If the node does not represent a parent folder, add its child nodes to the BFS queue to continue the traversal.
Step 9: After the BFS traversal is complete, return the `res` list, which contains the parent folder paths.

Key Insights

Insight 1: Sorting the folder paths by length allows us to efficiently check for sub-folders. A shorter path can only be a parent of a longer path, not the other way around.
Insight 2: Using a Trie data structure helps in quickly checking if a path is a sub-folder of any existing path.
Insight 3: Iterating through the folders and checking for sub-folders in a naive way (string prefix matching) can be inefficient; a Trie enables more efficient prefix checks.

Complexity Analysis

Time Complexity: O(n*l)

Space Complexity: O(n*l)

Topics

This problem involves: Array, String, Depth-First Search, Trie.

Companies

Asked at: Snowflake, Verkada.

Remove Sub-Folders from the Filesystem | LeetCode 1233

Medium

Array String Depth-First Search Trie

Time: O(n*l)Space: O(n*l)

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<String> removeSubfolders(String[] folder) {
        /**
         * Solves the LeetCode problem "Remove Sub-Folders from the Filesystem".
         *
         * This solution uses a Trie (prefix tree) to efficiently identify and
         * remove sub-folders.
         *
         * 1.  **Build a Trie:** Each folder path is inserted into a Trie. The Trie is
         *     implemented using nested HashMaps. Each part of a path (a directory
         *     name) is a key in a HashMap, and its value is another HashMap
         *     representing the next level. A special marker (e.g., a key '$') is
         *     used at the end of each path to store the full folder string.
         *
         * 2.  **Traverse the Trie:** A Breadth-First Search (BFS) is performed on the
         *     Trie, starting from the root.
         *
         * 3.  **Collect Parent Folders:** During the traversal, if a node contains the
         *     end-of-path marker, it signifies a parent folder. This folder's path
         *     is added to the result list. The traversal down this branch is then
         *     stopped, effectively pruning all its sub-folders.
         *
         * This method is optimal with a time and space complexity of O(N*L), where N
         * is the number of folders and L is the average length of a folder path.
         */
        Map<String, Object> trie = new HashMap<>();
        
        // 1. Build the Trie
        for (String path : folder) {
            Map<String, Object> node = trie;
            String[] parts = path.split("/");
            for (int i = 1; i < parts.length; i++) {
                String part = parts[i];
                if (!node.containsKey(part)) {
                    node.put(part, new HashMap<String, Object>());
                }
                node = (Map<String, Object>) node.get(part);
            }
            // Use a special marker to indicate the end of a path and store the path
            node.put("$", path);
        }
        
        List<String> res = new ArrayList<>();
        // 2. Traverse the Trie using BFS
        Queue<Map<String, Object>> q = new LinkedList<>();
        q.offer(trie);
        
        while (!q.isEmpty()) {
            Map<String, Object> node = q.poll();
            
            // 3. If a path ends here, it's a root folder. Add it to results.
            if (node.containsKey("$")) {
                res.add((String) node.get("$"));
                // Do not explore its children, as they are sub-folders
                continue;
            }
            
            // If not an end of a path, add children to the queue
            for (Object childNode : node.values()) {
                q.offer((Map<String, Object>) childNode);
            }
        }
        
        return res;
    }
}

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Problem Info

Description

Problem 1233: Remove Sub-Folders from the Filesystem - Detailed explanation and solution approaches.

Topics

Array String Depth-First Search Trie

Companies

SnowflakeVerkada

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1233

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Initialize an empty Trie (implemented using nested dictionaries or HashMaps).
2
Step 2: Iterate through the input `folder` list and insert each `path` into the Trie. Each level of the Trie represents a directory segment (e.g., 'a', 'b', 'c' in '/a/b/c'). A special marker ('$' in Python/Java or a boolean flag in C) is used to indicate the end of a path and store the full folder string at that node.
3
Step 3: Initialize an empty list `res` to store the parent folder paths.
4
Step 4: Perform a Breadth-First Search (BFS) on the Trie, starting from the root.
5
Step 5: In each BFS iteration, retrieve a node from the queue.
6
Step 6: Check if the current node contains the end-of-path marker. If it does, it means the node represents a parent folder.
7
Step 7: If the node represents a parent folder, add the corresponding folder path to the `res` list and skip exploring the children of that node. This effectively removes any sub-folders of the parent folder from consideration.
8
Step 8: If the node does not represent a parent folder, add its child nodes to the BFS queue to continue the traversal.
9
Step 9: After the BFS traversal is complete, return the `res` list, which contains the parent folder paths.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*l)

💾

Space Complexity

O(n*l)

📝

Why?

The time complexity is derived from the Trie construction and BFS traversal. Each operation involves iterating through folder paths and their components. The space complexity mirrors the Trie's potential size, influenced by the number of folders and their path lengths.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Sorting and Prefix Checking

Sort the folder paths by length in ascending order. Iterate through the sorted list and for each folder, check if any of the previously added folders are a prefix of the current folder. If not, add the current folder to the result. This method avoids the overhead of constructing the Trie.

Time: O(N*M*logN + N*M*M), where N is the number of folders and M is the average length of the folder paths. The N*M*logN comes from the sorting, and the N*M*M comes from the prefix checking (O(N) iterations to check prefix * M chars in prefix * O(M) to do comparison).Space: O(1) (excluding the space for the result list).

Trade-offs: This approach is simpler to implement, especially without prior knowledge of Trie. The primary solution using trie has better complexity when N is large. Sorting adds extra logarithmic complexity to the algorithm.

✨

Key Insights

1Insight 1: Sorting the folder paths by length allows us to efficiently check for sub-folders. A shorter path can only be a parent of a longer path, not the other way around.
2Insight 2: Using a Trie data structure helps in quickly checking if a path is a sub-folder of any existing path.
3Insight 3: Iterating through the folders and checking for sub-folders in a naive way (string prefix matching) can be inefficient; a Trie enables more efficient prefix checks.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty input list. The code should return an empty list in this case, which the given solution correctly handles.
Edge case 2: Input with only one folder. The code should return a list containing only that folder, which the given solution correctly handles.
Edge case 3: Overlapping folder names, such as '/a', '/a/b', '/a/bc'. The Trie structure handles this case correctly as the traversal will not stop at '/a' because it does not contain the end marker when '/a/bc' is being added.
Edge case 4: Identical folder paths, e.g., ['/a', '/a']. The Trie ensures only one instance of '/a' is stored.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly determining sub-folder relationships by not checking for the '/' after the potential parent folder's path. For example, '/a' should not be considered a parent of '/ab'.

✗Common mistake 2: Failing to handle edge cases such as an empty input list or an input list with only one folder.

✗Common mistake 3: Not understanding how the Trie's structure facilitates efficient prefix matching and attempting to implement prefix checks with string operations within loops, leading to higher time complexity.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: File system management. This algorithm can be used to identify and remove redundant sub-directories or to efficiently search for files within a directory structure.

🏦Real-world scenario 2: URL routing in web applications. Web frameworks often use a Trie-like structure to map URLs to handlers, and this algorithm can be used to resolve conflicts or identify more specific routes.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 1, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Remove Sub-Folders from the Filesystem"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?